Iklan

Iklan

Pertanyaan

x → 0 lim ​ tan 2 x sin 2 x cos 2 x − 1 ​ = ...

                    

  1.  negative 1    

  2. negative 0 comma 5        

  3. 0  

  4. 0 comma 5             

  5. 1  

Iklan

P. Anggrayni

Master Teacher

Jawaban terverifikasi

Jawaban

jawaban yang benar adalah B.

jawaban yang benar adalah B.

Iklan

Pembahasan

Ingatlah identitascosinus sudut rangkap serta sifat-sifat limit fungsi Dengan menggunakan sifat-sifat limit fungsi dan identitas tersebut, didapatkan : Maka, . Oleh karena itu, jawaban yang benar adalah B.

Ingatlah identitas cosinus sudut rangkap

cos space 2 x equals 1 minus 2 space sin squared space x

serta sifat-sifat limit fungsi

  • limit as x rightwards arrow c of k times f left parenthesis x right parenthesis equals k times limit as x rightwards arrow c of f left parenthesis x right parenthesis
     
  • limit as x rightwards arrow c of open square brackets f left parenthesis x right parenthesis times g left parenthesis x right parenthesis close square brackets equals limit as x rightwards arrow c of f left parenthesis x right parenthesis times limit as x rightwards arrow c of g left parenthesis x right parenthesis
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator tan space b x end fraction equals a over b
     
  • limit as x rightwards arrow 0 of fraction numerator sin space a x over denominator sin space b x end fraction equals a over b

Dengan menggunakan sifat-sifat limit fungsi dan identitas tersebut, didapatkan :

table attributes columnalign right center left columnspacing 0px end attributes row cell limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell equals cell limit as x rightwards arrow 0 of space fraction numerator open parentheses 1 minus 2 sin to the power of 2 space end exponent x close parentheses minus 1 over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell limit as x rightwards arrow 0 of space fraction numerator negative 2 sin to the power of 2 space end exponent x over denominator tan space 2 x space sin space 2 x end fraction end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of open parentheses fraction numerator sin space x over denominator tan space 2 x end fraction times fraction numerator sin space x over denominator sin space 2 x end fraction close parentheses end cell row blank equals cell negative 2 times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator tan space 2 x end fraction times limit as x rightwards arrow 0 of fraction numerator sin space x over denominator sin space 2 x end fraction end cell row blank equals cell negative 2 times 1 half times 1 half end cell row blank equals cell fraction numerator negative 2 over denominator 4 end fraction end cell row blank equals cell negative 0 comma 5 end cell end table   

Maka, limit as x rightwards arrow 0 of space fraction numerator cos space 2 x minus 1 over denominator tan space 2 x space sin space 2 x end fraction equals negative 0 comma 5

Oleh karena itu, jawaban yang benar adalah B.

Perdalam pemahamanmu bersama Master Teacher
di sesi Live Teaching, GRATIS!

2

Salsabilah Hasna

Bantu banget

Iklan

Iklan

Pertanyaan serupa

x → 0 lim ​ x tan x cos x − cos 2 x ​ = ...

207

5.0

Jawaban terverifikasi

RUANGGURU HQ

Jl. Dr. Saharjo No.161, Manggarai Selatan, Tebet, Kota Jakarta Selatan, Daerah Khusus Ibukota Jakarta 12860

Coba GRATIS Aplikasi Roboguru

Coba GRATIS Aplikasi Ruangguru

Download di Google PlayDownload di AppstoreDownload di App Gallery

Produk Ruangguru

Hubungi Kami

Ruangguru WhatsApp

+62 815-7441-0000

Email info@ruangguru.com

[email protected]

Contact 02140008000

02140008000

Ikuti Kami

©2024 Ruangguru. All Rights Reserved PT. Ruang Raya Indonesia